Does maximizing Jensen–Shannon divergence maximize Kullback–Leibler divergence? Does maximizing the Jensen–Shannon divergence $D_{\mathrm{JS}}(P \parallel Q)$ maximize the Kullback–Leibler divergence $D_{\mathrm{KL}}(P \parallel Q)$? If so, I'd like to be able to show that it does.
I have managed to express $D_{\mathrm{JS}}(P \parallel Q)$ in terms of $D_{\mathrm{KL}}(P \parallel Q)$ as follows:
$$D_{\mathrm{JS}}(P \parallel Q) = \frac{1}{2}D_{\mathrm{KL}}(P \parallel Q) + \frac{1}{2}D_{\mathrm{KL}}(Q \parallel P) - \frac{1}{2}\mathbb{E}_P\Big[ \log \Big(1 + \frac{P}{Q}\Big) \Big]  - \frac{1}{2}\mathbb{E}_Q\Big[ \log \Big(1 + \frac{Q}{P}\Big) \Big] + \log2$$
Does the linear relationship between the two in the above expression show that maximizing $D_{\mathrm{JS}}(P \parallel Q)$ also maximizes $D_{\mathrm{KL}}(P \parallel Q)$?
 A: I'm not sure whether the linear relationship plays a role, but I think you can show that if one attains its maximum then so does the other by the following reasoning (it is not a complete proof, I guess):
Both $D_{KL}$ and $D_{JS}$ are $f$-divergences, so we know that they attain their maximum when $P$ and $Q$ are orthogonal. Because $\max(D_{KL}) = \infty$,  $D_{KL}$ attains its maximum only if $P \perp Q$. The question remains whether $D_{JS}$ can attain its maximum $\log 2$ without $P \perp Q$.
Because $D_{KL}(P|Q)$, $D_{KL}(Q|P)$, $E_P[\log(1 +\frac{P}{Q})]$, and $E_Q[\log(1 +\frac{Q}{P})]$ are all positive, we can show that
$$ D_{KL}(P|Q) + D_{KL}(Q|P) \neq E_P[\log(1 +\frac{P}{Q})] + E_Q[\log(1 +\frac{Q}{P})] $$
by observing that
$$ \sum P \log \frac{P}{Q} + \sum Q \log \frac{Q}{P} < \sum P \log \frac{P+Q}{Q} + \sum Q \log \frac{P+Q}{P} $$
because
$$ \sum P \log \frac{P}{Q} < \sum P \log \frac{P+Q}{Q}$$
and are only equal for the limit when $P \perp Q$. And we conclude that if $D_{JS} = \log 2$, then $D_{KL} = \infty$.
A: Comment:
$D_{JS}(p \| q) =\frac{1}{2} D_{KL}(p \| m)+\frac{1}{2} D_{KL}(q \| m)$
$m=\frac{1}{2}(p+q)$
For the case where you assume Normal distribution for both $p \sim \mathcal N(\mu_1, \sigma_1)$ and $q \sim \mathcal N(\mu_2, \sigma_2)$, and $m \sim \mathcal N(\frac{\mu_1+\mu_2}{2}, \frac{\sigma_1 +\sigma_2}{2})$
You can express $D_{JS}$ in terms of $D_{KL}$
$D_{KL}(p \| q) = \log \frac{\sigma_{2}}{\sigma_{1}}+\frac{\sigma_{1}^{2}+\left(\mu_{1}-\mu_{2}\right)^{2}}{2 \sigma_{2}^{2}}-\frac{1}{2} $
and then it may be very easy to check.
